Trigo Identities!

**gp3** · February 18th 2010, 11:42 PM

hi guys,

solve cos2x - 3cosx +2 = 0 for 0 < x < 360

which formula should I use? is it the sum to product? I dont really know how to go about doing it! some advice pls!

thanks!

**Grandad** · February 19th 2010, 01:06 AM

Hello gp3

Originally Posted by gp3

hi guys,

solve cos2x - 3cosx +2 = 0 for 0 < x < 360

which formula should I use? is it the sum to product? I dont really know how to go about doing it! some advice pls!

thanks!

Use the identity:

$\cos 2x = 2\cos^2x-1$

Then simplify the LHS to get a quadratic in $\cos x$ , which you can then factorise.

Can you do it now?

Grandad

**gp3** · February 19th 2010, 01:39 AM

so what I've got now is:

2cos square x - 1 + 2 = 0
2cos square x + 1 = 0
(cosx+1)(cosx+1) = 0

is this correct?

**e^(i*pi)** · February 19th 2010, 01:45 AM

Originally Posted by gp3

so what I've got now is:

2cos square x - 1 + 2 = 0
2cos square x + 1 = 0
(cosx+1)(cosx+1) = 0

is this correct?

No, you've missed out your 3cos(x) term.
FYI $2cos^2(x)+1=0$ has no real solutions. This is because $x^2 \geq 0$ for all real x. Adding 1 will always result in a positive answer where x is real

The original question is $\cos(2x) - 3\cos(x) +2 = 0$

What Grandad said is to change $\cos(2x) to 2\cos^2(x)-1$ - the 3cos(x) term is unaffacted.

I have used square brackets to show this more effectively

$[2\cos^2(x) - 1] - 3cos(x) + 2 = 0$

You can simplify the constant terms to get it into the form $ax^2+bx+c=0$ :

$2\cos^2(x) - 3cos(x) + 1 = 0$

You can the solve the quadratic using your favourite method although this one factorises.

$(2\cos (x) -1)(\cos (x) - 1) = 0$

**gp3** · February 19th 2010, 01:49 AM

oh yea, thanks alot, I missed out 3cosx, guess I was too in a hurry.

and btw, just a quick one, how do you factorise the eqn, from 2cos square x -3 cosx+1+0 to (2 cosx-1)(cosx-1)= 0 ?

**Punch** · February 19th 2010, 03:06 AM

Originally Posted by gp3

oh yea, thanks alot, I missed out 3cosx, guess I was too in a hurry.

and btw, just a quick one, how do you factorise the eqn, from 2cos square x -3 cosx+1+0 to (2 cosx-1)(cosx-1)= 0 ?

Just solve it like any other quadratic equation. Treat cosx as x.

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